the tautological 1 form My question relates to p317-p318 of John Lee's "Introduction to Smooth Manifolds" discussion about the tautological 1 form.
In Proposition 12.24, we have the expression:
$\tau_{(x, \xi)} = \pi^* (\xi_i dx^i) = \xi_i dx^i  $
Here $\tau_{(x, \xi)} \in T^*_{(x, \xi)}(T^*Q)$ and $\xi_i dx^i \in T^*_x Q$.  Does this mean $\xi_i dx^i$ is in both $ T^*_x Q$ and $T^*_{(x, \xi)}(T^*Q)$ ?
In addition, I have some follow up questions because I'm a bit confused about the notation:
If we have a point $p$ on a manifold $M$, and a tangent vector $v \in T_pM$, its standard coordinate for $TM$ is given by the pair $(p, v)$.  What is its standard coordinate in $TTM$?  is it still $(p, v)$?
Similarly, for a point $q \in Q$ on the smooth manifold $Q$ with its covector $\varphi \in T^*_q Q$, we have the standard coordinates for $T^*Q$ being $(q, \varphi)$.  Then are the standard coordinates on $T^*(T^*Q)$ still $(q, \varphi)$ ?
 A: It is true that $\xi_i \, dx^i$ is in both $T_x^\ast Q$ and $T^\ast_{(x,\xi)} T^\ast Q$. Locally, in a neighborhood $U$ of $(x, \xi) \in T^\ast Q$, you have bundle charts of the form $U \times \Bbb R^{2n}$ for $T^\ast (T^\ast Q)$. The coordinates in the $U$ part of the chart will just be the coordinates $(x_1, \dots, x_n, \xi_1, \dots, \xi_n)$ in $T^\ast Q$. There will be an extra $2n$ coordinates in the "fiber direction" $\Bbb R^{2n}$ for $T^\ast (T^\ast Q)$, however. Therefore the local coordinates for $T^\ast Q$ and $T^\ast (T^\ast Q)$ are not exactly the same ($T^\ast (T^\ast Q)$ has more coordinates).
The same thing happens even with just the cotangent bundle $T^\ast Q$ (or any bundle at all over $Q$). We can take a small enough neighborhood $U \subset Q$ with coordinates $(x_1, \dots, x_n)$, and there will be a corresponding chart for $T^\ast Q$ of the form $U \times \Bbb R^n$ with coordinates $(x_1, \cdots, x_n, \xi_1, \dots, \xi_n)$. The $x_i$ coordinates are in both $Q$ and $T^\ast Q$, but the $\xi_i$ coordinates (in the "fiber direction") are not in $Q$.
